Simple Complex Vector Space Proof Clarification

[RJLiberator]

Homework Statement

Let v ∈ V and c ∈ ℂ, with c ≠ 0. Prove that if cv = 0, then v = 0.

Homework Equations

Vector space axioms.

The Attempt at a Solution

Simple proof overall, but I have one major clarification question.

v = 1v
= (c^(-1)c)v
= c^(-1) (cv)
= c^(-1) 0
v = 0

My question is, in a complex vector space, is it safe to assume that c^(-1) exists in this proof?
If it does, I feel very confident about this proof.
If it doesn't then I need to do something else.

I don't see anywhere in the vector space axioms that states the multiplicative inverse exists.
But, it makes sense to me that an inverse does exist for any possible scalar choice here.

 

[S.G. Janssens]

But, it makes sense to me that an inverse does exist for any possible scalar choice here.

Yes indeed, because ##\mathbb{C}## is a field. More generally, one always speaks of a vector space over a field. (Real and complex numbers are the common choices, but there are others.) So the existence of ##\mathbb{c}^{-1}## for ##\mathbb{c} \neq 0## really follows from the field axioms, not from the vector space axioms.

 

[RJLiberator]

Beautiful! I was searching the vector space axioms and had them in my head. The field axioms clearly state this.

thank you, once again, Krylov.